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Today I was fiddling about with a TI-89 calculator, attempting as usual to confuse it. I figured that making it solve an equation with a periodic function would be fun, so I tried the following:

$$\cos(\pi x) + x^2 = 0$$

While this didn't stump it, I noted that the solution it gave was just a decimal. Since I ran this through solve(), which usually will give you a nice solution like $\frac{\sqrt{2}}{2}$ if it can, I found this rather interesting. I figured what it eventually did was give up on solving it in its usual manner and move to a numeric method: i.e., it simply made intelligent guesses until it found good solutions.

I am usually inclined to think most random decimals have a closed form expression behind them. (In fact, do all such decimals have a closed form, even if we don't know it? I may have to look into that.) As such, I decided to put this into Wolfram|Alpha and see if it had any better results. But no dice; it gave me back, similar to the TI-89, that $x = \pm 1$ and $x \approx \pm 0.629847$ were solutions.

The latter decimal, $x \approx 0.629847$, is the one I am concerned with. As far as I am concerned, $x = \pm 1$ are sort of "trivial" solutions; just thinking the problem over leads you to them naturally.

Is there a way to solve this algebraically? I can sort of narrow it down. I know that as $x \to \infty$, the $\cos(\pi x)$ term is essentially trivial compared to $x^2$. Given that $x^2>0$ for any $x \ne 0$, and given that $\cos(\pi x)$ has a range of $[-1, 1]$, it seems to me that whenever $x^2>1$, $\cos(\pi x)$ cannot pull down $x^2$ enough for it become zero.

So, it seems natural to me to think, then, that all solutions must lie where $x^2 \le 1$, viz., within the interval $[-1, 1]$. This narrows down the field significantly, but it still does not really help me with an algebraic solution. (However, I figure if I were to come across this in a real-world scenario, this would be a useful line of attack for a guess-and-test sort of deal.)

Another line of attack I attempted was to take the reverse approach: use the numerically attained solution to find a closed-form solution. I thought it may be an interesting number I simply had not learned about, so I tried to look up the decimal sequence in the OEIS, but to no luck: no such sequence was available.

I've sort of rambled, so here are my questions:

  1. Is there an algebraic solution to the above equation?
  2. Even if there is not, is there any way to figure out the closed form expression behind the decimal $x = 0.629847$? I don't even care if the expression has $\cos$ or $\sin$ in it.

I will be honest: I really don't even know where to start.

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5  
(1) Highly doubtful. (2) Every finite decimal has the closed-form of a rational number, but "almost all" infinite decimals have no closed form. This has to do with cardinality: the set of all definable numbers will be countably infinite while the real numbers are uncountably many. –  anon May 17 '12 at 3:25
3  
As a nice, general, rule-of-thumb: any transcendental equation where the variable you're solving for appears within the transcendental function (in your case, the cosine) and outside the function, will generally not have a simple closed-form solution. –  J. M. May 17 '12 at 3:27
    
Expressing cosine in terms of exponentials, can give $(-1)^{x+1} = x^2.$ (I might've made a mistake, though.) –  user2468 May 17 '12 at 3:36
2  
Closed form, in general, is a very deceptive term. What do you mean by it? If you get a solution say $\pi/4$, will you call it closed form? If so, then I can define, $\Upsilon$ is the real number number between $0$ and $1$ such that $\cos(\pi \Upsilon) + \Upsilon^2 = 0$. And now lo and behold, your $x = \Upsilon$. Would you call this closed form? It really doesn't mean much if you think about it. –  user17762 May 17 '12 at 6:57

2 Answers 2

up vote 9 down vote accepted

Your $x \approx 0.629847$ is transcendental. First, $x \neq 0, \; 1/3, \; 2/3, \; 1 \;$ so by Corollary 3.12 in Niven, Irrational Numbers, page 41, $x$ is irrational. Now, one value of $$ (-1)^x $$ is $$ e^{i \pi x} = \cos \pi x + i \sin \pi x. $$ As $x$ is not rational, Gelfond-Schneider, Theorem 10.1 on page 134, $ (-1)^x $ is transcendental.

The algebraic numbers in $\mathbb C$ are a field containing $i,$ the rationals, and closed under complex conjugation. It follows that a number is algebraic if and only if both its real and imaginary parts are algebraic. From $\cos^2 \pi x + \sin^2 \pi x = 1,$ it follows that either both parts are algebraic or both are transcendental. Therefore $\cos \pi x$ is transcendental. Since $x^2 = - \cos \pi x,$ we find that $x$ itself is transcendental.

Meanwhile, lots of nice numbers are transcendental. $\pi, \; e, \; \log 2 \; $ are transcendental but would be considered pleasant answers to a problem of this type. It is very, very hard to show that a number such as your $x$ does not have a nice closed form expression. I can't see how it could, of course.

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The inverse symbolic calculator gives a few dozen numbers starting with .629847; maybe if you could get a couple more decimal places you'd narrow it down. Of course, Will is right that it's unlikely to have a simple expression in terms of familiar constants such as $\pi$, $\sqrt2$, etc., but it still might be a value of some special function someone has tabulated.

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Thanks for the tool. Unfortunately, while the results look promising for the first few digits, even expanding to the next "section" of digits with Wolfram|Alpha ends up with a decimal that's not in that tool's database. –  Reid May 17 '12 at 13:41
1  
Well, I'm not all that surprised, but I figured it was worth a try - and now you know the "calculator" is there, maybe it will come in handy in the future. –  Gerry Myerson May 17 '12 at 13:46

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